A) \[\frac{mgR}{Bl}\]
B) \[\frac{mgR}{{{B}^{2}}{{l}^{2}}}\]
C) \[\frac{mgR}{{{B}^{3}}{{l}^{3}}}\]
D) \[\frac{mgR}{{{B}^{2}}l}\]
Correct Answer: B
Solution :
Due to magnetic field, wire will experience an upward force \[F=Bil=B\ \left( \frac{Bvl}{R} \right)\ l\]\[\Rightarrow F=\frac{{{B}^{2}}v{{l}^{2}}}{R}\] If wire slides down with constant velocity then \[F=mg\Rightarrow \frac{{{B}^{2}}v{{l}^{2}}}{R}=mg\Rightarrow v=\frac{mgR}{{{B}^{2}}{{l}^{2}}}\]
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